CHM 2210 Final Exam V3 | CHM 2210
Organic Chemistry I | Actual Q&A with
Rationale (CHM2210 Final Exam) |
University of Central Florida
1. Which of the following statements best describes the hybridization and geometry of the
nitrogen atom in ammonia (NH3)?
A. sp2, trigonal planar
B. sp2, bent
C. sp3, tetrahedral
D. sp3, trigonal pyramidal
Answer: D
Rationale: The nitrogen atom in ammonia has three bonding pairs and one lone pair of
electrons. According to VSEPR theory, these four electron domains require sp3
hybridization to minimize repulsion. The presence of the lone pair reduces the bond angles
and results in a trigonal pyramidal molecular geometry.
2. Identify the strongest Brønsted-Lowry acid among the following compounds.
A. Ethanol (CH3CH2OH)
B. Water (H2O)
C. Acetic acid (CH3COOH)
,D. Trichloroacetic acid (Cl3CCOOH)
Answer: D
Rationale: The acidity of a carboxylic acid is significantly enhanced by the presence of
electron-withdrawing groups. In trichloroacetic acid, the three chlorine atoms exert a
powerful inductive effect that stabilizes the conjugate base. This stabilization makes it
easier for the molecule to donate a proton compared to acetic acid or alcohols.
3. Which of the following alkyl halides will react fastest in an SN2 reaction with sodium
ethoxide?
A. 2-iodo-2-methylpropane
B. 2-iodobutane
C. 1-chlorobutane
D. 1-iodobutane
Answer: D
Rationale: SN2 reactions are highly sensitive to steric hindrance at the electrophilic center.
Primary alkyl halides react much faster than secondary or tertiary ones due to easier
nucleophilic attack. Additionally, iodine is a better leaving group than chlorine because it is
a weaker base and more polarizable.
4. What is the relationship between (2R,3R)-2,3-dibromobutane and (2S,3S)-2,3-
dibromobutane?
A. Enantiomers
, B. Identical compounds
C. Diastereomers
D. Constitutional isomers
Answer: A
Rationale: Enantiomers are non-superimposable mirror images of each other. In this case,
every chiral center has its configuration inverted from R to S. Since the molecules are
mirror images and lack an internal plane of symmetry, they are classified as enantiomers.
5. Which reagent is most suitable for the anti-Markovnikov hydration of an alkene?
A. BH3-THF followed by H2O2, NaOH
B. Hg(OAc)2, H2O followed by NaBH4
C. OsO4, NMO
D. H2O, H2SO4
Answer: A
Rationale: Hydroboration-oxidation is a two-step sequence that adds water across a
double bond with anti-Markovnikov regioselectivity. The boron atom adds to the less
substituted carbon due to steric and electronic factors. Subsequent oxidation replaces the
boron with a hydroxyl group, retaining the regiochemical outcome.
Organic Chemistry I | Actual Q&A with
Rationale (CHM2210 Final Exam) |
University of Central Florida
1. Which of the following statements best describes the hybridization and geometry of the
nitrogen atom in ammonia (NH3)?
A. sp2, trigonal planar
B. sp2, bent
C. sp3, tetrahedral
D. sp3, trigonal pyramidal
Answer: D
Rationale: The nitrogen atom in ammonia has three bonding pairs and one lone pair of
electrons. According to VSEPR theory, these four electron domains require sp3
hybridization to minimize repulsion. The presence of the lone pair reduces the bond angles
and results in a trigonal pyramidal molecular geometry.
2. Identify the strongest Brønsted-Lowry acid among the following compounds.
A. Ethanol (CH3CH2OH)
B. Water (H2O)
C. Acetic acid (CH3COOH)
,D. Trichloroacetic acid (Cl3CCOOH)
Answer: D
Rationale: The acidity of a carboxylic acid is significantly enhanced by the presence of
electron-withdrawing groups. In trichloroacetic acid, the three chlorine atoms exert a
powerful inductive effect that stabilizes the conjugate base. This stabilization makes it
easier for the molecule to donate a proton compared to acetic acid or alcohols.
3. Which of the following alkyl halides will react fastest in an SN2 reaction with sodium
ethoxide?
A. 2-iodo-2-methylpropane
B. 2-iodobutane
C. 1-chlorobutane
D. 1-iodobutane
Answer: D
Rationale: SN2 reactions are highly sensitive to steric hindrance at the electrophilic center.
Primary alkyl halides react much faster than secondary or tertiary ones due to easier
nucleophilic attack. Additionally, iodine is a better leaving group than chlorine because it is
a weaker base and more polarizable.
4. What is the relationship between (2R,3R)-2,3-dibromobutane and (2S,3S)-2,3-
dibromobutane?
A. Enantiomers
, B. Identical compounds
C. Diastereomers
D. Constitutional isomers
Answer: A
Rationale: Enantiomers are non-superimposable mirror images of each other. In this case,
every chiral center has its configuration inverted from R to S. Since the molecules are
mirror images and lack an internal plane of symmetry, they are classified as enantiomers.
5. Which reagent is most suitable for the anti-Markovnikov hydration of an alkene?
A. BH3-THF followed by H2O2, NaOH
B. Hg(OAc)2, H2O followed by NaBH4
C. OsO4, NMO
D. H2O, H2SO4
Answer: A
Rationale: Hydroboration-oxidation is a two-step sequence that adds water across a
double bond with anti-Markovnikov regioselectivity. The boron atom adds to the less
substituted carbon due to steric and electronic factors. Subsequent oxidation replaces the
boron with a hydroxyl group, retaining the regiochemical outcome.
